if x + 1/x = 5
then, find out the value of x and then find out x^2 + 1/x^2 =?
F2 maths question
2010-06-12 6:59 pm
回答 (4)
2010-06-12 7:02 pm
✔ 最佳答案
x + 1/x = 5(x + 1/x)^2 = 5^2
x^2 + 2x(1/x) + (1/x)^2 = 25
x^2 + 2 + 1/x^2 = 25
x^2 + 1/x^2 = 23
2010-06-12 11:07:21 補充：
value of x :
x + 1/x = 5
x^2 + 1 = 5x
x^2 - 5x + 1 = 0
x = {-(-5) +/- √[5^2 - 4(1)(1)] } / [2(1)]
x = (5 +/- √21) / 2
2010-06-13 7:51 pm
take square from (1)
(x+1/x)^2=25
x^2+2+1/x^2=25
x^2+1/x^2=24
from 1
-x(x-5)=1
-x^2+5x=1
5x-1=x^2
take (4)into (3) then
5x-1+1/(5x-1)=24
25x^2+10x+1=120x-24
25x^2=110x-25
25x-110=-25x
50x-110=0
x=-(110-50)
=2.2
(x+1/x)^2=25
x^2+2+1/x^2=25
x^2+1/x^2=24
from 1
-x(x-5)=1
-x^2+5x=1
5x-1=x^2
take (4)into (3) then
5x-1+1/(5x-1)=24
25x^2+10x+1=120x-24
25x^2=110x-25
25x-110=-25x
50x-110=0
x=-(110-50)
=2.2
2010-06-13 12:31 am
x + (1/x) = 5
x^2-2(1/x)(x)+(1/x^2)=25
x^2+2+(1/x^2)=25
x^2+(1/x^2)=23
x^2-2(1/x)(x)+(1/x^2)=25
x^2+2+(1/x^2)=25
x^2+(1/x^2)=23
2010-06-12 7:07 pm
(x+1/x)=5
(x+1/x)^2=5^2<----其實都係欠2次方，左右兩邊2次方試吓la
x^2+2(x)(1/x)+(1/x^2)=25
x^2+2+1/x^2=25
x^2+1/x^2=25-2
x^2+1/x^2=23
(x+1/x)^2=5^2<----其實都係欠2次方，左右兩邊2次方試吓la
x^2+2(x)(1/x)+(1/x^2)=25
x^2+2+1/x^2=25
x^2+1/x^2=25-2
x^2+1/x^2=23
收錄日期: 2021-04-21 22:15:48
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https://hk.answers.yahoo.com/question/index?qid=20100612000051KK00411